A small bulb (assumed to be a point source) is placed at the bottom of a tank containing water to a depth of `80 cm`. Find out the area of the surface of water through which light from the bulb can emerge. Take the value of refractive index of water to be `4//3`.

Explanation

In Fig. the source of light (S) is `80cm` below the surface of water i.e., `SO = 80 cm = 0.8 m`.
When `angle i = C, for SA and SB, angle r = 90^(@)`
`:.` Area of the surface of water through which light from the bulb can emerge is area of the circle of radius
`r = (AB)/(2) = OA = OB`
As `mu = (1)/(sin C)`
`sin C = (1)/(mu) = (1)/(1.33) = 0.75`
`C = sin^(-1) (0.75) = 48.6^(@)`
In `DeltaOBS`, `tan C = (OB)/(OS) :. OB = OS tan C = 0.8 tan 48.6^(@)`
Area of the surface of water through which light emerges `= pir^(2) = 3.14 (0.907)^(2) = 2.518 m^(2)`